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[?] Aharonov–Bohm Effect and Toroidal Transformers

  1. Aug 11, 2012 #1
    There is probably a very simple and obvious answer to this question, but right now it's just not clicking for me. So, perhaps someone else can help me out.

    Context:

    The typical undergrad intro to the Aharonov–Bohm effect is that it is an intrinsically non-classical phenomenon because an electron is effected by travelling through a region where E=0 and B=0. The claim is that, classically, regions with A≠0 but B=E=0 cannot have any effect on particles. Hence the 'quantumness' of the Aharonov–Bohm effect.

    So then, what about a toroidal transformer? (i.e. a toroidal loop of iron with a powered wire coil on one side and an unpowered wire coil on the other) We know that even if the coils don't touch the iron torus that the powered one can still induce a current in the unpowered one. However, unless I'm mistaken, you can construct such a device for which B=0 outside the iron. Yet the transformer still operates.

    Even more generally, Faraday's law says that an EMF will be induced in a loop of wire that encloses a changing magnetic field. As far as I know, it doesn't say anything about the B field having to be nonzero in the region where the wire loop is. One could imagine creating a confined changing B field, whether by using a superconducting shell or by using a Aharonov–Bohm type field configuration, and then surrounding it with a wire coil (that is, so that the coil is outside of the confinement region). Faraday's law would imply that an EMF will be induced in the wire, no?

    The question:

    Considering these two similar classical situations, how can one say that regions of A≠0 but B=E=0 produce no classical effect? It seems to me that electrons are affected by such regions in the case of classical induction.
     
  2. jcsd
  3. Aug 11, 2012 #2

    tom.stoer

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    You are right, a changing flux can induce a current, but:
    1) there is no changing flux in the Aharonov-Bohm, setup; all fields are constant
    2) microscopically in classical electrodynamics the force acting on the electrons (which is responsible for inducing a current) is the Lorentz force v x B; so if B=0 the force must vanish
     
  4. Aug 11, 2012 #3
    Absolutely. I don't mean to say that the two situations are completely analogous.

    EDIT: I guess I should clarify since I mentioned "or by using a Aharonov–Bohm type field configuration". What I meant was a variation of the Aharonov–Bohm setup where there is a *time-varying* B field that is only non-zero in a confined cylinder. Certainly this is different from the Aharonov–Bohm setup. Fore example, one might consider the case of the Aharonov–Bohm setup where the solenoid is turned on and off cyclically. I don't see any reason why it should be impossible to reproduce the Aharonov–Bohm type configuration with the modification of a time-varying field.

    So, are you then saying that if you have a confined, time-varying B field that it would not induce a current in any loop circling it outside the confinement boundary? I thought this happened in toroidal transformers. If I am mistaken and the confinement does make a difference, then Faraday's law is quite misleading when stated in the conventional way, because the traditional statement only refers to the time changing flux that is enclosed by the loop and has no concern for whether or not the B field is non-zero at the location of the loop itself.
     
    Last edited: Aug 11, 2012
  5. Aug 11, 2012 #4
    I think you need to read maxwell equation for help. You see, E is induced at the the postion where B changes.
     
  6. Aug 11, 2012 #5
    I think the answer might just be that Faraday's law, in the form of

    EMF=-N dΦ/dt

    just isn't always true in general. My point is that law doesn't seem to care whether the "enclosed flux" is non-zero at the actual location of the wire. It only cares what the total time changing flux is, where the flux is the total integral of B * dA in the entire enclosed region. Clearly you can have a wire enclosing a time varying Φ without the wire passing through a region where B≠0.

    However, I believe it is already well established that Faraday's law cannot be naively applied to every situation, and that there are limitations on it. Perhaps this is such an example?

    I think I am confusing the issue by talking too much. For the sake of clarity, let's forget Aharonov–Bohm for a moment and just focus on this one question:

    Will an EMF be induced in a wire coil that surrounds a confined, time-varying B field?

    Here is a possible setup for an undergrad homework-type question I found while googling the issue: http://www.chegg.com/homework-help/...olume-radius-006-m-b-directed-plane--q1243598


    EDIT: I think I have the answer. Courtesy of page 10-11 of http://ocw.mit.edu/courses/physics/...-fall-2010/faradays-law/MIT8_02SC_notes21.pdf which pointed out an enormous oversight I was making.

    Yes a wire enclosing a confined B field will have an induced EMF and current, even if the wire is in a region where B=0. The reason is because E≠0.

    The solution lies in the fact that the above quote is false! Although maxwell's equations relate the curl of E and the derivative of B at the same point, an E field can be induced even in regions where B is constant and 0!.

    So then the final answer is that in the classical cases of toroidal transformers and confined time-varying B fields, the electrons are in regions where B=0 but E≠0 in those regions due to the time varying B, and hence there is motion. Whereas in the Aharonov–Bohm experiment, E is forced to be 0 as well simply because B is static.
     
    Last edited: Aug 11, 2012
  7. Aug 12, 2012 #6

    tom.stoer

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    Yes, the confusion goes away once you are using the differential form of Maxwell's equations. Then it becomes clear that currents (of charges) can only be induced in regions where charges do exist, i.e. not in vacuum, and where the B-field is non-vanishing.

    Other statements regarding changing E-fields but vanishing B-field are confusing b/c according to Maxwell's equations you cannot have B=0 but non-vanishing and changing (!) E-fields in a certain region of space. Vanishing B-fields allow only constant E-fields. But constant E-fields cannot induce a current in a ring b/c constant E-fields are gradient fields.

    In addition the Aharanov-Bohm effect does not violate these equations b/c it does not imply a current or a force.
     
  8. Aug 12, 2012 #7
    Yes,I have to say I made a simple mistake. E can be nonzero even its curl is 0. But I think the question is clear now. Your question's condition that both E and B both equal 0 can not be satisfied. So there should not be a question"how can one say that regions of A≠0 but B=E=0 produce no classical effect?"

    But what do you mean by saying ' Faraday's law cannot be naively applied to every situation, and that there are limitations on it.' Of course Faraday's law is right!
     
  9. Aug 12, 2012 #8
    Fair enough, but the Aharanov–Bohm effect isn't even statable in terms of classical E&M. Moreover, we are talking about forcefully shooting single electrons at a screen to produce interference; so, I don't even think it's meaningful to speak of whether or not the effect violates Maxwell's equations. Nevertheless, I get your point. (Although I'm sure you'd find a violation in the simple fact that an interference pattern forms. That's the whole magic about the effect; that it doesn't happen classically. But I digress.)

    The only reason I brought Aharonov–Bohm into this is because I was trying to point out a situation where charges are classically affected by a region where A≠0 and E=B=0. I think we can all agree that this issue is settled simply because the scenarios I came up with had B=0 in the region of interesest but E≠0. So I had erred in my assumptions.

    Now, with that issue settled, there remains an entirely separate, classical issue: 'Will a current be induced in a wire loop that encloses a confined time-varying magnetic field even if B=0 in the region of the wire?'


    You say the answer is no:
    Here we disagree.

    Here we agree. And indeed the situation I am referring to has a constant E field in the region of B=0. In fact, E is constant everywhere in the scenario.

    Here is the heart of our disagreement. It is for a subtle reason. Look at the setup on page 10-11 of http://ocw.mit.edu/courses/physics/...-fall-2010/faradays-law/MIT8_02SC_notes21.pdf again. E is constant. But E is certainly not conservative. Try a line integral ∮E∙ds around a circular path (centered about z=0) that encloses the magnetic region but is outside the confinement region. The answer is obviously nonzero, without even doing the math. (The field is clearly macroscopically rotational, and it is constant along circles centered about z=0.) Although, if one insists on doing the math anyway the answer will be 2πR² dB/dt (big R, not little r). It should be clear then that there will be a current induced in any circular loop, even if r>R.

    So E is NOT conservative. The definition of conservative (up to equivalency) is that all closed line integrals are zero. The confusion is in the fact that the curl vanishes, and we are used to saying '0 curl implies conservative field' due to Stoke's theorem. But,[STRIKE] first of all,[/STRIKE] that is only true if the curl is 0 everywhere. In this case the curl is 0 in the region of the line integral, but there are regions where the curl is nonzero. (In fact, our integral encloses such a region). [STRIKE]Secondly, there is a cusp at r=R, so we can't appeal to Stoke's theorem anyway.[/STRIKE] See "Aside" below.

    The fact that E is not conservative shouldn't be surprising though, because it is well known that conservative changing B fields can create nonconservative E fields. What might be surprising is that conservative changing B fields can create E fields that are nonconservative even when they are only examined in regions for which B=0, but as the above example shows it is true.





    Aside:
    Remember, the E field in this scenario is piecewise defined. But even if we setup the constant E field E=R²/2r dB/dt everywhere in space (so it is not piecewise defined and the curl is 0 'everywhere'), it still wouldn't be conservative and there'd still be a current. Namely, because the curl is 0 almost everywhere, but not everywhere. There is a singularity at r=0. So such a field isn't physical anyway.


    EDIT: Striked out the bits about the cusp as you end up appealing to Stoke's theorem to solve for E (depending on what form of maxwell's equations you use). Really, the cusp bit is only an issue because of the discontinuity in B, and I don't imagine that it fowls anything up besides some of the limiting details at that exact spot. I'm sure everything is a lot more rigorous if you just pick a smoother function for B that drops off continuously to 0.
     
    Last edited: Aug 12, 2012
  10. Aug 12, 2012 #9
    Exactly.

    Well it turns out that Faraday's law does work here, because there is an EMF and it is given by the same old formula. I was merely looking for ways in which the situation could get away with not inducing an EMF even though Faraday's law says 'it should regardless of whether or not B=0 in the region of the wire'. However, in general, I believe that quote is valid and that there are funky situations that defy a "naive" application of Faraday's law. I was referring to http://en.wikipedia.org/wiki/Faraday's_law_of_induction#.22Counterexamples.22_to_Faraday.27s_law which draws from the Feynman lectures. However, as that link points out, you can use Faraday's law if you are very careful about your path of integration.
     
  11. Aug 13, 2012 #10

    tom.stoer

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    -dove, please have a look at Maxwell's equations!

    They say

    [tex]\nabla \times E + \frac{\partial B}{\partial t} = 0[/tex]

    So a non-vanishing curl of E is incompatible with zero and constant (!) B. And vice versa it means that only a time-dependent B can induce a non-zero curl of E. But a time-dependent B is incompatible with zero B. So constant E with non-vanishing curl E but zero and constant B is incompatible with Maxwell's equations.

    When I am saying that "the Aharanov-Bohm effect does not violate these equations b/c it does not imply a current or a force. " then I mean exactly what I have written down: The Aharanov-Bohm effect does not violate Maxwell's equations. They are still valid even when the electron is treated via Schrödinger's equation. There is no current induced on the qm level! This can be stated explicitly in the qm framework.

    Then please think about "... that currents (of charges) can only be induced…where the B-field is non-vanishing. " Please look at Ampere's law (in differential form - up to some constants).

    [tex]\nabla \times B - \frac{\partial E}{\partial t} = j[/tex]

    It says that a current requires a combination of curl of B and time-dependent E; but b/c a time-dep. E always induces a B-field (as we have seen above) you will never get a current with static fields !! (unless you move the wire through a static B-field which is irrelevant here b/c we do not move any wire and we do not have such a B-field).

    Your question 'Will a current be induced in a wire loop that encloses a confined time-varying magnetic field even if B=0 in the region of the wire?' does not make much sense. As long as the B-field is still zero in the region of the wire it will not induce a current b/c there is no B-field (Ampere's law). But a time-varying B-field produces el.-mag. waves! They will arrive at the wire and will induce a current, but then you don't have a vanishing B-field anymore.

    To shorten the discussion: can you please tell me how exactly the field configuration shall look like? In terms of B(x,t) and E(x,t)?
     
    Last edited: Aug 13, 2012
  12. Aug 13, 2012 #11
    Yes. Like I said, I get your point. Maxwell's equations continue to be valid for the E and B (or ɸ and A), of course. My point was just that one should be careful not to then get the idea that classical electrodynamics gives the right answer. The classical failing is a result of the qE+q(v×B)=F=ma picture being mistaken, it is not a failing of Maxwell's equations, and clearly you understand that. I just mean that one should be careful with such statements in a Quantum Physics help forum lest someone become confused thinking that the Aharonov–Bohm effect doesn't violate classical E&M.


    [Emphasis Added.]

    I am well aware! It is the region in bold that highlights our miscommunication.

    We are clearly miscommunicating. I thought I had made the field configuration clear, but I think it got lost in a lot of nonsense. Please look at the following link that I posted earlier: http://www.chegg.com/homework-help/...olume-radius-006-m-b-directed-plane--q1243598

    The situation is very simple, as is the analysis. Take r to be the cylindrical radius coordinate and take R to be some fixed value. You setup up a B field that is nonzero and time-varying in some region r≤R. Let's take dB/dt to be constant and positive in this central region. The B field is confined to this region so B=0 for r>R. Then you wrap a wire loop around the B field region in a circle whose radius is greater than R. At this precise moment, the B field in the wire is 0.

    A current will be induced in the wire. The reason being that the central B field we supplied induces an E field that circles around and around, as shown on page 10-11 of http://ocw.mit.edu/courses/physics/...-fall-2010/faradays-law/MIT8_02SC_notes21.pdf. This E field is nonzero even in the region where B=0. The E field is constant and zero curl, compatible with Maxwell's equation, but it still produces an EMF due to a nonzero closed line integral. The EMF produced is consistent with Faraday's law.

    So, even though the wire was placed in a region where B=0, a current was still induced, namely because the B field from the center region had induced an E field in the region where the wire was placed. Yes, as soon as the current starts flowing there will be a B field at the point of the wire and a time-varying E field.

    Hopefully I have cleared up our miscommunication.



    The whole thing has absolutely nothing to do with Aharonov–Bohm, and like I suspected, I quickly realized that I was being silly. I had forgotten to calculate the induced E field. The fact that we don't supply an E field means squat. It was early for me, and I hadn't slept much :tongue:
     
    Last edited: Aug 13, 2012
  13. Aug 13, 2012 #12

    tom.stoer

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    It seems to me that the arguments presentetd in the paper are inconsistent.

    On page 10 they say "let’s consider a uniform magnetic field ... confined to a circular region with radius R".

    But on page 12 they write down an equation

    [tex]E_{nc} = \frac{R^2}{2r}\frac{dB}{dt}[/tex]

    Now
    a) either B is confined to r<R, then B=0 for r>R and E=0; or
    b) E ≠ 0 (as is shown in figure 10.3.2), then dB/dt ~ rE and therefore B ≠ 0 for r>R

    You cannot have both a confined B vanishing outside R and a non-zero E.

    The following paper may help: http://www.slac.stanford.edu/cgi-wrap/getdoc/slac-pub-3514.pdf
     
    Last edited: Aug 13, 2012
  14. Aug 13, 2012 #13
    What they mean by dB/dt is the constant value it takes in the confined region. They do not mean the value it takes at the point at which you evaluate E. This is evident if you see how they derive that formula. E~1/r outside R and E~r inside R. The proportionality constant for the latter case is R^2 times the proportionality constant for the former. In both cases dB/dt is just a label for a specific value. E is nonzero everywhere except for along the z axis.

    In regions of vanishing B, the curl of the electric field must be 0. This does not imply that the field itself must be 0. I am not sure what further restraints you are appealing to in order to conclude that a vanishing B implies a vanishing E.

    Thanks for the paper, it is very interesting. You can always rely on Griffiths for highlighting important, but subtle mathematical details. However, I am not sure what relevance it has to the configuration we are discussing. They are concerned with fields generated by certain current flows. Here we are talking about setting up a particular B field (independent of the details of how one would actually make such a configuration) and we are asking what the E field must be to be consistent with Maxwell's equations.

    For the slightly different case of toroidal transformers (which I mentioned earlier), a user named "pam" hits the nail on the head in the following thread: https://www.physicsforums.com/showthread.php?t=212410.
    Furthermore, the user "dot" who posted that thread made exactly the same oversight that I did when I made this thread. The B field vanishes outside toroidal transformers, but the E field does not. There is nothing spooky about it, and it is not an example where particles can be affected by regions of E=B=0 (unlike Aharonov–Bohm).
     
    Last edited: Aug 13, 2012
  15. Aug 13, 2012 #14

    tom.stoer

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    The paper still confuses me but we should go ahead w/o it

    Please correct me I am wrong.

    You start with a solenoid with radius R and B(r) = 0 for r>R. We don't care for the inner solution with time-dependent B-field at the moment. From Maxwell's equations we get

    [tex]\nabla \times E + \frac{\partial B}{\partial t} = 0\qquad\stackrel{B=0}{\Rightarrow}\qquad \nabla \times E = 0[/tex]

    [tex]\nabla \times B - \frac{\partial E}{\partial t} = j\qquad\stackrel{B=0}{\Rightarrow}\qquad - \frac{\partial E}{\partial t} = j[/tex]

    b/c the current j is zero, E must not be time-dependent.

    Do you agree on this?

    Do you agree that if we try to introduce a time-dependent E-field but still insist on B=0 we automatically have a non-vanishing current which is forbidden for r>R? Therefore the only allowed solution is E=const.

    Now E can be written as

    [tex]E = -\nabla \phi - \frac{\partial A}{\partial t}[/tex]

    with curl-free A (b/c otherwise the curl of A would result in a non-vanishing B)

    An explicit solution is given as (in our reference)

    [tex]E_\text{out} = \frac{aR^2}{2r}e_\phi[/tex]

    In my last post I made the wrong conclusion that vanishing (confined) B-field implies a vanishing E-field. That was wrong and we can agree on the static E-field, can't we?

    Now we have a time-independent E-field with vanishing curl but non-vanishing line integral around a closed loop with fixed R'>R.

    What would be your next step?
     
    Last edited: Aug 14, 2012
  16. Aug 14, 2012 #15
    All looks good to me!

    Now, with this field configuration a priori, all I am saying is that if you then wrap a wire around to make a circle centered about the z axis, and this wire loop is entirely in the region of vanishing B, you will still find an EMF and current induced in this wire. Of course, once the current begins to flow, it will change the E and B fields in that region: i.e. the B field will no longer be vanishing.
     
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